Algebraic Topology Disjoint Union, Lee

Question

As this is my first post, I hope that what I write is pretty clear and isn't too disappointing. Without further ado, I am quite curious with regards to my understanding of Disjoint Union Spaces. (I am currently self studying from Introduction to Topological Manifolds by Lee.)

First, I will begin with a definition.

Def: Suppose $(X_a)_{a\in A}$ is an indexed family of non-empty topological spaces. Denote their disjoint union by $\coprod_{a\in A}X_a$.

We define the $\textbf{canonical injection}$ by: $\iota_a: X_a\rightarrow \coprod_aX_a$ with the assignment being $\iota_a(x)=(x,a)$.

As a matter of convention, we force the equality: $X_a= \iota_a(X_a)$.

We define define a topology on $\coprod_aX_a$ by declaring that $U\subseteq \coprod_aX_a$ is open if and only if $U\cap X_a$ is open in $X_a$ for each a.

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Lee, in his book, states that $X_a$ in $U\cap X_a$ is considered as a subset of the disjoint union. Does this mean, ignoring the equality convention above, that he really means $\iota_a^{-1}(U)$ is open in $X_a$, for each a?

Secondly, what is the purpose of the convention above?. Lastly, what are good ways to start thinking about the disjoint union?

Answer 1

In general, let $X$ be a set and $(X_i, \tau_i)_{i\in I}$ a family of topological spaces. Every family $\{f_i: X_i\longrightarrow X\}_{i\in I}$ of mappings induce a topology on $X$ called the final topology with respect to this family.

By definition, the final topology on $X$ with respect to the family $\{f_i: X_i\longrightarrow X\}_{i\in I}$ is the finest topology on $X$ which turns each $f_i: X_i\longrightarrow X$ into a continuous map.

It turns out, $U$ is open in this topology if and only if $f_i^{-1}(U)$ is open in $X_i$ for every $i\in I$.

The disjoint union topology is a special case, can you see why?

See final topology for further details.

You may also want to check the dual notion called initial topology.

As to the convention, notice

$$\imath_a: X_a\longrightarrow \imath_a(X_a)$$ are bijections. In those cases, it is usual to identity the domain with its image to simplify the notations.

Finally, you should think of the disjoint union as a way to force the members of the union to be disjoint, since:

$$(i, X_i)=(j, X_j)\Leftrightarrow i=j.$$ Think of every element of the indexing set $A$ as a point in the plane and sitting on the point $i\in A$ there is the set $X_i$.